1) \( \left(\frac{2}{3} x^{2}-\frac{3}{5} y^{3}\right)^{3}= \)

To expand the expression \( \left(\frac{2}{3} x^{2}-\frac{3}{5} y^{3}\right)^{3} \), we can use the binomial theorem, which states that \( (a - b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^{k} \). Here, let \( a = \frac{2}{3} x^{2} \), \( b = \frac{3}{5} y^{3} \), and \( n = 3 \). Applying the theorem, we get: \[ \left(\frac{2}{3} x^{2}-\frac{3}{5} y^{3}\right)^{3} = \sum_{k=0}^{3} \binom{3}{k} \left(\frac{2}{3} x^{2}\right)^{3-k} \left(-\frac{3}{5} y^{3}\right)^{k} \] Calculating the individual terms results in: 1. For \( k = 0 \): \( \binom{3}{0} \left(\frac{2}{3} x^{2}\right)^{3} = \left(\frac{2}{3}\right)^{3} x^{6} = \frac{8}{27} x^{6} \) 2. For \( k = 1 \): \( \binom{3}{1} \left(\frac{2}{3} x^{2}\right)^{2} \left(-\frac{3}{5} y^{3}\right) = 3 \cdot \frac{4}{9} x^{4} \cdot \left(-\frac{3}{5} y^{3}\right) = -\frac{36}{45} x^{4} y^{3} = -\frac{4}{5} x^{4} y^{3} \) 3. For \( k = 2 \): \( \binom{3}{2} \left(\frac{2}{3} x^{2}\right)^{1} \left(-\frac{3}{5} y^{3}\right)^{2} = 3 \cdot \frac{2}{3} x^{2} \cdot \frac{9}{25} y^{6} = \frac{54}{75} x^{2} y^{6} = \frac{18}{25} x^{2} y^{6} \) 4. For \( k = 3 \): \( \binom{3}{3} \left(-\frac{3}{5} y^{3}\right)^{3} = \left(-\frac{3}{5}\right)^{3} y^{9} = -\frac{27}{125} y^{9} \) Combining these results, we have: \[ \left(\frac{2}{3} x^{2}-\frac{3}{5} y^{3}\right)^{3} = \frac{8}{27} x^{6} - \frac{4}{5} x^{4} y^{3} + \frac{18}{25} x^{2} y^{6} - \frac{27}{125} y^{9} \] So, the final expanded expression is: \[ \frac{8}{27} x^{6} - \frac{4}{5} x^{4} y^{3} + \frac{18}{25} x^{2} y^{6} - \frac{27}{125} y^{9} \]